HOC24
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Chủ đề / Chương
Bài học
\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
\(=\left(x^2-1\right)\left(x+2\right)\)
\(=x^3+2x^2-x-2\)
Do x = 14
=> 15 = x + 1 ; 16 = x + 2 ; 29 = 2x + 1 ; 13 = x - 1
\(\Rightarrow B=x^5-\left(x+1\right)x^4+\left(x+2\right)x^3-\left(2x+1\right)x^2+\left(x-1\right)x\)
\(\Rightarrow B=x^5-x^5-x^4+x^4+2x^3-2x^3-x^3+x^2-x\)
\(\Rightarrow B=-x=-14\)
\(x^5+x-1\)
\(=x^5+x^2-x^2+x-1\)
\(=x^2\left(x^3+1\right)-\left(x^2-x+1\right)\)
\(=x^2\left(x+1\right)\left(x^2-x+1\right)-\left(x^2-x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^3+x^2-1\right)\)
\(Q=\dfrac{x^2-3x+3}{\left(x-1\right)^2}\)
\(Q=\dfrac{x^2-2.\dfrac{3}{2}x+\dfrac{9}{4}+\dfrac{3}{4}}{\left(x-1\right)^2}\)
\(Q=\dfrac{\left(x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}}{\left(x-1\right)^2}\)
Để Q có GTNN
=> \(\left(x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\)phải lớn nhất và (x-1)^2 phải bé nhất
\(2x^2+2y^2-2xy-4x-4y+8\)
\(=x^2-2xy+y^2+x^2-4x+y^2-4y+8\)
\(=\left(x-y\right)^2+x^2-4x+4+y^2-4x+4\)
\(=\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2\ge0\)
\(\RightarrowĐPCM\)
\(\dfrac{x}{2}+\dfrac{x}{y}-\dfrac{3}{2}=\dfrac{10}{y}\)
\(\dfrac{x-3}{2}=\dfrac{10-x}{y}\)
\(\Rightarrow\left(x-3\right)y=2\left(10-x\right)\)
\(\Rightarrow xy-3y-20+2x=0\)
\(\Rightarrow x\left(y+2\right)-3\left(y+2\right)-14=0\)
\(\Rightarrow\left(x-3\right)\left(y+2\right)=14\)
Lập bảng là ra nha bn
\(\left(a+b+c\right)^2=\dfrac{9}{4}\)
\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=\dfrac{9}{4}\)
Có \(a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)
\(b^2+c^2\ge2\sqrt{b^2c^2}=2bc\)
\(a^2+c^2\ge2\sqrt{a^2c^2}=2ac\)
\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc\le a^2+b^2+c^2+a^2+b^2+a^2+c^2+b^2+c^2=3\left(a^2+b^2+c^2\right)\)
\(\Rightarrow\dfrac{9}{4}\le3\left(a^2+b^2+c^2\right)\)
\(\Rightarrow a^2+b^2+c^2\ge\dfrac{9}{4}.\dfrac{1}{3}=\dfrac{3}{4}\left(ĐPCM\right)\)
Bài này áp dụng BĐT cosi nha bn
\(VP=a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(VP=\left(a+b\right)\left(a^2-ab+b^2\right)+c^3+3\left(a +b\right)\left(a+c\right)\left(b+c\right)\)
\(=\left(a+b\right)\left(a^2+2ab+b^2-3ab\right)+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)
\(=\left(a+b\right)^3+c^3-3\left(a+b\right)\left[\left(ab+ac+bc+c^2\right)\right]\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-ac-bc+c^2\right]-3\left(a+b\right)\left(a+b\right)\left(a+c\right)\)
\(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
Làm tương tự
\(=\dfrac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}=\dfrac{5^{32}-1}{2}\)